Tip:
Highlight text to annotate it
X
So in terms of an example, let's talk about a very common interference anti-reflective coating
Normally because it is wavelength dependent, we normally target the middle of the visible spectrum
so that we can get that effect
on average for the entire spectrum, so it works ideally
centered at green light whose wavelength in air is 550 nm
by depositing a layer of MgF2 which has a certain index, on glass
with a certain other index, and we are asking what is the thickness we need to achieve
the anti-reflective effect
let's draw this out
we have a piece of glass here, n = 1.52 down here
and we deposit some material on top
and the n is 1.38 in here, and this is going to have to work in air, so the n out here is 1
green light comes in
and bounce off, we have our first beam
bounce off the second interface, second beam
we're talking about
anti-reflections, so we want destructive
so that's our +/-π, +/-3π, ...
and we want to figure the total phase, we have to
first of all, treat and make sure the reflection at each point, do we get a phase shift
for number 1
φ1 here
that's going to equal to π because you have
going from n of 1 to n of 1.38, Low to High, π
same thing over here
φ2
1.38 to 1.52
so that's also low to high, π
therefore, the entire Δφ going on here is
...
and k here of course is not just
the simple k, it's going to be ...
times the index of refraction
of the thin film itself, because that's the part where the extra path was taken over, so that goes in there
and Δx is going to be two times the thickness which is what we are looking for
so the minimum we can get
for destructive is then, we will put in π, that's the smallest
we can get, not 3π, not 5π; -π wouldn't work so it has to be π
rearranging we get that
...
...
you can think of it as the wavelength in the medium
whoops, over 4 because it's 2 x 2
So it's a quarter of the wavelength, but because it is corrected by the index of refraction, it can be even thinner
working out that number, we get 99.6 nm
and that's the thickness of the anti-reflective coating to minimize reflection at 550 nm
now we should be concerned though, because it is
wavelength dependent, just because it is destructive at a certain wavelength, that should mean that
it is constructive for other wavelengths, so we have to make sure that it is not going to be bother us either
so to make that
constructive
for constructive, we know the thickness this time, we want to know what λ we are working with this time
but for constructive, we are going to use 2π
and so that's ...
....
...
...
... in air
now this kind of wavelength is deep within the purple towards ultraviolet where our eyes aren't that sensitive
so even though it is amplifying the reflection at that frequency, it doesn't hurt us as bad
we are minimizing the frequency where our eyes are sensitive in the greens
and we are going to have a little more reflection where our eyes aren't that sensitive
and that's OK, it achieves the effect of an anti- reflective coating
so once again, to recap, as you do these problems
usually it is a good idea to sketch it out to know which n goes where
then you can work out the phase shift for each individual reflection
write out this equation
don't forget your n and then work out the answer